Is Permittivity Equivalent to EM Conductivity?

[Dickfore]

So in this case we get division by zero?

Provided that $\hat{\sigma}(\omega) \neq o(\omega), \ \omega \rightarrow 0$. This is true for conductors, but not for insulators (dielectrics).

EDIT:
Remember, we are working wIth the Fourier transforms of these quantities in the frequency domain. This simply tells us that there ought to be a pole of the dielectric response function at $\omega = 0$. Going back to time domain, we get:
$$\epsilon''(t -t') = \frac{\sigma}{\epsilon_0} \, \theta(t - t')$$
which gives the following relation between the polarization and the electric field:
$$\mathbf{P}(t) = \sigma \, \int_{-\infty}^{t}{\mathbf{E}(t') \, dt'}$$
or, the current density due to bound charges is:
$$\mathbf{J}(t) = \dot{\mathbf{P}}(t) = \sigma \, \mathbf{E}(t)$$
But, this is just Ohm's law in differential form, provided the "bound" charges are the free ones, as is the case for conductors.